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| | + | Note: Please do not post problems here until after the AIME. |
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| | == Problem == | | == Problem == |
| − | Let <math>ABCD</math> be a [[parallelogram]]. Extend <math>\overline{DA}</math> through <math>A</math> to a point <math>P,</math> and let <math>\overline{PC}</math> meet <math>\overline{AB}</math> at <math>Q</math> and <math>\overline{DB}</math> at <math>R.</math> Given that <math>PQ = 735</math> and <math>QR = 112,</math> find <math>RC.</math>
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| | == Solution == | | == Solution == |
| − | === Solution 1 ===
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| − | [[Image:AIME_1998-6.png|350px]]
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| − | There are several [[similar triangles]]. <math>\triangle PAQ\sim \triangle PDC</math>, so we can write the [[proportion]]:
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| − | <div style="text-align:center;">
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| − | <math>\frac{AQ}{CD} = \frac{PQ}{PC} = \frac{735}{112 + 735 + RC} = \frac{735}{847 + RC}</math>
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| − | </div>
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| − | Also, <math>\triangle BRQ\sim DRC</math>, so:
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| − | <div style="text-align:center;">
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| − | <math>\frac{QR}{RC} = \frac{QB}{CD} = \frac{112}{RC} = \frac{CD - AQ}{CD} = 1 - \frac{AQ}{CD}</math><br />
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| − | <math>\frac{AQ}{CD} = 1 - \frac{112}{RC} = \frac{RC - 112}{RC}</math>
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| − | </div>
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| − | Substituting,
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| − | <div style="text-align:center;">
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| − | <math>\frac{AQ}{CD} = \frac{735}{847 + RC} = \frac{RC - 112}{RC}</math><br />
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| − | <math>735RC = (RC + 17797)(RC - 13)</math><br />
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| − | <math>0 = RC^2 - 13\cdot17797</math>
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| − | </div>
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| − | Thus, <math> RC = \sqrt{13*17797} = 481</math>.
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| − | === Solution 2 ===
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| − | We have <math>\triangle BRQ\sim \triangle DRC</math> so <math>\frac{112}{RC} = \frac{BR}{DR}</math>. We also have <math>\triangle BRC \sim \triangle DRP</math> so <math>\frac{ RC}{847} = \frac {BR}{DR}</math>. Equating the two results gives <math>\frac{13}{RC} = \frac{ RC}{17797}</math> and so <math>RC^2=13*17797</math> which solves to <math>RC=\boxed{481}</math>
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Revision as of 17:40, 27 February 2020
Note: Please do not post problems here until after the AIME.
Problem
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